很CONCEPT的COMPLEX NO.

1. 設θ= 2π/7, z^7 - 1 = 0
=> (z-1)(z^2- 2zcosθ+1)(z^2- 2zcos2θ+1)(z^2- 2zcos3θ+1)=0
2. 預備公式
(1) A= cosθ+cos(2θ)+cos(3θ) = - 1/2
(2) B= cosθcos(2θ)+cosθcos(3θ)+ cos(2θ)cos(3θ)= -1/2
(3) C= cosθcos(2θ)cos(3θ) = 1/8
3. 設 x= (z^2 +1)/(2z) 代入
(z^2- 2zcosθ+1)(z^2- 2zcos2θ+1)(z^2- 2zcos3θ+1)= 0
=> (x - cosθ)(x-cos2θ)(x-cos3θ)=0
=> x^3 - A x^2+Bx - C =0
=> 8x^3+ 4x^2 - 4x -1 = 0
=> x=(z^2 + 1)/(2z) , z= cos(kθ) + i sin(kθ) , k=1,2,3,4,5,6
=> x= cos(kθ), k=1,2,3
so, x=cos(2π/7), cos(4π/7), cos(6π/7)

HOW DO YOU GET 2

(1) The factorization of z^7 - 1 should be
(z - 1)(z^2 - 2zcos2π/7 + 1)(z^2 - 2zcos4π/7 + 1)(z^2 - 2zcos6π/7 + 1) instead.

(2) Do you mean "to obtain a cubic equation whose roots are cos2π/7, cos4π/7 and cos6π/7"?